chapter 8: matrices and determinants - math notes and math tests

(Section 8.1: Matrices and Determinants) 8.01

CHAPTER 8: MATRICES and DETERMINANTS

The material in this chapter will be covered in your Linear Algebra class (Math 254 at Mesa).

SECTION 8.1: MATRICES and SYSTEMS OF EQUATIONS

PART A: MATRICES

A matrix is basically an organized box (or “array”) of numbers (or other expressions).In this chapter, we will typically assume that our matrices contain only numbers.

Example

Here is a matrix of size 2 × 3 (“2 by 3”), because it has 2 rows and 3 columns:

1 0 2

0 1 5

⎡

⎣⎢

⎤

⎦⎥

The matrix consists of 6 entries or elements.

In general, an m × n matrix has m rows and n columns and has mn entries.

Example

Here is a matrix of size 2 × 2 (an order 2 square matrix):

4 −1

3 2

⎡

⎣⎢

⎤

⎦⎥

The boldfaced entries lie on the main diagonal of the matrix.(The other diagonal is the skew diagonal.)


PART B: THE AUGMENTED MATRIX FOR A SYSTEM OF LINEAR EQUATIONS

Example

Write the augmented matrix for the system:

3x + 2y + z = 0

−2x − z = 3

⎧⎨⎩

Solution

Preliminaries:

Make sure that the equations are in (what we refer to now as)standard form, meaning that …

• All of the variable terms are on the left side (with x, y, and zordered alphabetically), and

• There is only one constant term, and it is on the right side.

Line up like terms vertically.

Here, we will rewrite the system as follows:

3x + 2y + z = 0

−2x − z = 3

⎧⎨⎩

(Optional) Insert “1”s and “0”s to clarify coefficients.

3x + 2y +1z = 0

−2x + 0y −1z = 3

⎧⎨⎩

Warning: Although this step is not necessary, people oftenmistake the coefficients on the z terms for “0”s.


Write the augmented matrix:

Coefficients of Right x y z sides

3 2 1

−2 0 −1

0

3

⎡

⎣⎢

⎤

⎦⎥

Coefficient matrix Right-hand side (RHS)

Augmented matrix

We may refer to the first three columns as the x-column, they-column, and the z-column of the coefficient matrix.

Warning: If you do not insert “1”s and “0”s, you may want to read theequations and fill out the matrix row by row in order to minimize thechance of errors. Otherwise, it may be faster to fill it out column bycolumn.

The augmented matrix is an efficient representation of a system oflinear equations, although the names of the variables are hidden.


PART C: ELEMENTARY ROW OPERATIONS (EROs)

Recall from Algebra I that equivalent equations have the same solution set.

Example

Solve: 2x −1= 5

2x −1= 5

2x = 6

x = 3 ⇒ Solution set is 3{ }.

To solve the first equation, we write a sequence of equivalent equations untilwe arrive at an equation whose solution set is obvious.

The steps of adding 1 to both sides of the first equation and of dividing bothsides of the second equation by 2 are like “legal chess moves” that allowedus to maintain equivalence (i.e., to preserve the solution set).

Similarly, equivalent systems have the same solution set.

Elementary Row Operations (EROs) represent the legal moves that allow us to write asequence of row-equivalent matrices (corresponding to equivalent systems) until weobtain one whose corresponding solution set is easy to find. There are three types ofEROs:


1) Row Reordering

Example

Consider the system:

3x − y = 1

x + y = 4

⎧⎨⎩

If we switch (i.e., interchange) the two equations, then the solution setis not disturbed:

x + y = 4

3x − y = 1

⎧⎨⎩

This suggests that, when we solve a system using augmented matrices,…

We can switch any two rows.

Before:

R1

R2

3 −1

1 1

1

4

⎡

⎣⎢

⎤

⎦⎥

Here, we switch rows R1 and R2

, which we denote

by: R1↔ R

2

After:

new R1

new R2

1 1

3 −1

4

1

⎡

⎣⎢

⎤

⎦⎥

In general, we can reorder the rows of an augmented matrixin any order.

Warning: Do not reorder columns; in the coefficient matrix,that will change the order of the corresponding variables.


2) Row Rescaling

Example


1

2x +

1

2y = 3

y = 4

⎧⎨⎪

⎩⎪

If we multiply “through” both sides of the first equation by 2, then weobtain an equivalent equation and, overall, an equivalent system:

x + y = 6

y = 4

⎧⎨⎩

This suggests that, when we solve a system using augmented matrices,…

We can multiply (or divide) “through” a row by anynonzero constant.

Before:

R1

R2

1 / 2 1 / 2

0 1

3

4

⎡

⎣⎢

⎤

⎦⎥

Here, we multiply through R1 by 2, which we

denote by: R1← 2 ⋅ R

1, or

new R

1( )← 2 ⋅ old R1( )

After:

new R1

R2

1 1

0 1

6

4

⎡

⎣⎢

⎤

⎦⎥


3) Row Replacement

(This is perhaps poorly named, since ERO types 1 and 2 may also be viewedas “row replacements” in a literal sense.)

When we solve a system using augmented matrices, …

We can add a multiple of one row to another row.

Technical Note: This combines ideas from the Row Rescaling EROand the Addition Method from Chapter 7.

Example


x + 3y = 3

−2x + 5y = 16

⎧⎨⎩

Before:

R1

R2

1 3

-2 5

3

16

⎡

⎣⎢

⎤

⎦⎥

Note: We will sometimes boldface items for purposes of clarity.

It turns out that we want to add twice the first row to the secondrow, because we want to replace the “ -2 ” with a “0.”

We denote this by:

R2← R

2+ 2 ⋅ R

1, or

new R

2( )← old R2( ) + 2 ⋅ R

1

old R2 − 2 5 16

+2 ⋅ R1 2 6 6

new R2 0 11 22


Warning: It is highly advised that you write out the table!People often rush through this step and make mechanical errors.

Warning: Although we can also subtract a multiple of one rowfrom another row, we generally prefer to add, instead, even ifthat means that we multiply “through” a row by a negativenumber. Errors are common when people subtract.

After:

old R1

new R2

1 3

0 11

3

22

⎡

⎣⎢

⎤

⎦⎥

Note: In principle, you could replace the old R1 with the

rescaled version, but it turns out that we like having that “1” inthe upper left hand corner!

If matrix B is obtained from matrix A after applying one or more EROs, then wecall A and B row-equivalent matrices, and we write A B .

Example

1 2

7 8

3

9

⎡

⎣⎢

⎤

⎦⎥

7 8

1 2

9

3

⎡

⎣⎢

⎤

⎦⎥

Row-equivalent augmented matrices correspond to equivalent systems, assumingthat the underlying variables (corresponding to the columns of the coefficientmatrix) stay the same and are in the same order.


PART D: GAUSSIAN ELIMINATION (WITH BACK-SUBSTITUTION)

This is a method for solving systems of linear equations.

Historical Note: This method was popularized by the great mathematician Carl Gauss,but the Chinese were using it as early as 200 BC.

Steps

Given a square system (i.e., a system of n linear equations in n unknowns for some

n ∈Z+ ; we will consider other cases later) …

1) Write the augmented matrix.

2) Use EROs to write a sequence of row-equivalent matrices until you get one inthe form:

If we begin with a square system, then all of the coefficient matrices will besquare.

We want “1”s along the main diagonal and “0”s all below.The other entries are “wild cards” that can potentially be any real numbers.

This is the form that we are aiming for. Think of this as “checkmate” or“the top of the jigsaw puzzle box” or “the TARGET” (like in a trig ID).

Warning: As you perform EROs and this form crystallizes and emerges,you usually want to avoid “undoing” the good work you have already done.For example, if you get a “1” in the upper left corner, you usually want topreserve it. For this reason, it is often a good strategy to “correct” thecolumns from left to right (that is, from the leftmost column to therightmost column) in the coefficient matrix. Different strategies may workbetter under different circumstances.


For now, assume that we have succeeded in obtaining this form; thismeans that the system has exactly one solution.

What if it is impossible for us to obtain this form? We shall discuss thismatter later (starting with Notes 8.21).

3) Write the new system, complete with variables.

This system will be equivalent to the given system, meaning that they sharethe same solution set. The new system should be easy to solve if you …

4) Use back-substitution to find the values of the unknowns.

We will discuss this later.

5) Write the solution as an ordered n-tuple (pair, triple, etc.).

6) Check the solution in the given system. (Optional)

Warning: This check will not capture other solutions if there are, in fact,infinitely many solutions.

Technical Note: This method actually works with complex numbers in general.

Warning: You may want to quickly check each of your steps before proceeding. A singlemistake can have massive consequences that are difficult to correct.


Example

Solve the system:

4x − y = 13

x − 2y = 5

⎧⎨⎩

Solution

Step 1) Write the augmented matrix.

You may first want to insert “1”s and “0”s where appropriate.

4x − 1y = 13

1x − 2y = 5

⎧⎨⎩

R1

R2

4 −1

1 −2

13

5

⎡

⎣⎢

⎤

⎦⎥

Note: It’s up to you if you want to write the “ R1” and the “ R2

.”

Step 2) Use EROs until we obtain the desired form:

1 ?

0 1

?

?

⎡

⎣⎢

⎤

⎦⎥

Note: There may be different “good” ways to achieve our goal.

We want a “1” to replace the “4” in the upper left.Dividing through R1

by 4 will do it, but we will then end up with

fractions. Sometimes, we can’t avoid fractions. Here, we can.

Instead, let’s switch the rows.

R1↔ R

2

Warning: You should keep a record of your EROs. This will reduceeyestrain and frustration if you want to check your work!

R1

R2

1 −2

4 −1

5

13

⎡

⎣⎢

⎤

⎦⎥


We now want a “0” to replace the “4” in the bottom left.Remember, we generally want to “correct” columns from left to right,so we will attack the position containing the −1 later.

We cannot multiply through a row by 0.

Instead, we will use a row replacement ERO that exploits the “1” inthe upper left to “kill off” the “4.” This really represents theelimination of the x term in what is now the second equation in oursystem.

new R

2( )← old R2( ) + −4( ) ⋅ R

1

The notation above is really unnecessary if you show the work below:

old R2 4 − 1 13

+ −4( ) ⋅ R

1 − 4 8 − 20

new R2 0 7 -7

R1

R2

1 −2

0 7

5

−7

⎡

⎣⎢

⎤

⎦⎥

We want a “1” to replace the “7.”We will divide through R2

by 7, or, equivalently, we will multiply

through R2 by

1

7:

R

2←

1

7⋅ R

2, or

R1

R2

1 −2

0 7

5

−7

⎡

⎣⎢

⎤

⎦⎥

← ÷7

R1

R2

1 −2

0 1

5

−1

⎡

⎣⎢

⎤

⎦⎥


We now have our desired form.

Technical Note: What’s best for computation by hand may not be bestfor computer algorithms that attempt to maximize precision andaccuracy. For example, the strategy of partial pivoting would havekept the “4” in the upper left position of the original matrix and wouldhave used it to eliminate the “1” below.

Note: Some books remove the requirement that the entries along themain diagonal all have to be “1”s. However, when we refer toGaussian Elimination, we will require that they all be “1”s.

Step 3) Write the new system.

You may want to write down the variables on top of theircorresponding columns.

x y

1 −2

0 1

5

−1

⎡

⎣⎢

⎤

⎦⎥

x − 2y = 5

y = −1

⎧⎨⎩

↑

This is called an upper triangular system, which is very easy to solveif we …


Step 4) Use back-substitution.

We start at the bottom, where we immediately find that y = -1 .

We then work our way up the system, plugging in values forunknowns along the way whenever we know them.

x − 2y = 5

x − 2 −1( ) = 5

x + 2 = 5

x = 3

Step 5) Write the solution.

The solution set is:

3, -1( ){ }Books are often content with omitting the { } brace symbols.

Ask your instructor, though.

Warning: Observe that the order of the coordinates is the reverse ofthe order in which we found them in the back-substitution procedure.

Step 6) Check. (Optional)

Given system:

4x − y = 13

x − 2y = 5

⎧⎨⎩

4 3( ) − −1( ) = 13

3( ) − 2 −1( ) = 5

⎧⎨⎪

⎩⎪

13= 13

5 = 5

⎧⎨⎩

Our solution checks out.


Example (#62 on p.556)

Solve the system:

2x + 2y − z = 2

x −3y + z = −28

− x + y = 14

⎧

⎨⎪

⎩⎪

Solution

Step 1) Write the augmented matrix.

You may first want to insert “1”s and “0”s where appropriate.

2x + 2y − 1z = 2

1x − 3y + 1z = −28

−1x + 1y + 0z = 14

⎧

⎨⎪

⎩⎪

R1

R2

R3

2 2 −1

1 −3 1

−1 1 0

2

−28

14

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Step 2) Use EROs until we obtain the desired form:

1 ? ?

0 1 ?

0 0 1

?

?

?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

We want a “1” to replace the “2” in the upper left corner.Dividing through R1

by 2 would do it, but we would then end up

with a fraction.

Instead, let’s switch the first two rows.

R1↔ R

2


R1

R2

R3

1 −3 1

2 2 −1

-1 1 0

−28

2

14

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

We now want to “eliminate down” the first column by using the “1”in the upper left corner to “kill off” the boldfaced entries and turnthem into “0”s.

Warning: Performing more than one ERO before writing down a newmatrix often risks mechanical errors. However, when eliminatingdown a column, we can usually perform several row replacementEROs without confusion before writing a new matrix. (The same istrue of multiple row rescalings and of row reorderings, which canrepresent multiple row interchanges.) Mixing ERO types beforewriting a new matrix is probably a bad idea, though!

old R2 2 2 − 1 2

+ −2( ) ⋅ R

1 − 2 6 − 2 56

new R2 0 8 -3 58

old R3 − 1 1 0 14

+R1 1 − 3 1 − 28

new R3 0 -2 1 -14

Now, write down the new matrix:

R1

R2

R3

1 −3 1

0 8 −3

0 −2 1

−28

58

−14

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

The first column has been “corrected.” From a strategic perspective,we may now think of the first row and the first column (in blue) as“locked in.” (EROs that change the entries therein are not necessarily“wrong,” but you may be in danger of being taken further away fromthe desired form.)


We will now focus on the second column. We want:

1 −3 1

0 1 ?

0 0 ?

−28

?

?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Here is our current matrix:

R1

R2

R3

1 −3 1

0 8 −3

0 -2 1

−28

58

−14

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

If we use the “ -2 ” to kill off the “8,” we can avoid fractions for the

time being. Let’s first switch R2 and R3

so that we don’t get confused

when we do this. (We’re used to eliminating down a column.)

Technical Note: The computer-based strategy of partial pivotingwould use the “8” to kill off the “ -2 ,” since the “8” is larger inabsolute value.

R2↔ R

3

R1

R2

R3

1 −3 1

0 -2 1

0 8 −3

−28

−14

58

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Now, we will use a row replacement ERO to eliminate the “8.”

old R3 0 8 − 3 58

+4 ⋅ R2 0 − 8 4 − 56

new R3 0 0 1 2

Warning: Don’t ignore the “0”s on the left; otherwise, you may getconfused.


Now, write down the new matrix:

R1

R2

R3

1 −3 1

0 -2 1

0 0 1

−28

−14

2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Once we get a “1” where the “ -2 ” is, we’ll have our desired form.We are fortunate that we already have a “1” at the bottom of the thirdcolumn, so we won’t have to “correct” it.

We will divide through R2 by −2 , or, equivalently, we will multiply

through R2 by

−

1

2.

R

2← −

1

2

⎛⎝⎜

⎞⎠⎟⋅ R

2, or

R1

R2

R3

1 −3 1

0 -2 1

0 0 1

−28

−14

2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

← ÷ −2( )

We finally obtain a matrix in our desired form:

R1

R2

R3

1 −3 1

0 1 −1 / 2

0 0 1

−28

7

2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥


Step 3) Write the new system.

x y z

1 −3 1

0 1 −1 / 2

0 0 1

−28

7

2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

x − 3y + z = −28

y −1

2z = 7

z = 2

⎧

⎨⎪⎪

⎩⎪⎪

↑↑

Step 4) Use back-substitution.

We immediately have: z = 2

Use z = 2 in the second equation:

y −1

2z = 7

y −1

22( ) = 7

y −1= 7

y = 8

Use y = 8 and z = 2 in the first equation:

x − 3y+ z = −28

x − 3 8( )+ 2( ) = −28

x − 24 + 2 = −28

x − 22 = −28

x = -6


Step 5) Write the solution.

The solution set is:

-6, 8, 2( ){ }Warning: Remember that the order of the coordinates is the reverse ofthe order in which we found them in the back-substitution procedure.

Step 6) Check. (Optional)

Given system:

2x + 2y − z = 2

x −3y + z = −28

− x + y = 14

⎧

⎨⎪

⎩⎪

2 −6( ) + 2 8( ) − 2( ) = 2

−6( ) − 3 8( ) + 2( ) = −28

− −6( ) + 8( ) = 14

⎧

⎨⎪⎪

⎩⎪⎪

2 = 2

−28 = −28

14 = 14

⎧

⎨⎪

⎩⎪

Our solution checks out.

chapter 8: matrices and determinants - math notes and math tests

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